Question

plz zzz help me...... ​

Answer 1

${\purple{ \large\sf \: Newton's \: 2nd \: Law \: of \: Motion}}$

According to Newton's 2nd law of motion The rate of change of momentum of a body is directly proportional to the applied force,and takes place in the direction in which the force acts. The rate of change of momentum of a body can be obtained by dividing the Change in momentum by Time taken for change. So,It can be expressed by :

$\sf \: Force \propto \dfrac{Change \: in \: momentum}{Time \: taken}$

Consider a body of Mass m having an initial velocity u. The initial momentum of this body will be mu. Suppose a force F acts on this body for time t & causes the final velocity to become v. The final momentum of this body will be mv. Now,the change in momentum of this body is mv - mu & the time taken for this change is t. So, According to Newton's First Law of Motion :

$\large \: \sf \: F \propto \dfrac{ mv \: - \: mu}{t}$

or

$\large \: \sf \: F \propto \dfrac{ m(v - u)}{t}$

But $\sf\frac{v-u}{t}$ represents change in velocity with time which is known as acceleration a. So,by writing a in place of $\sf\frac{v-u}{t}$ in the above relation,we get :

$\large \: \sf \: F \propto m \times a$

Thus, the force acting on a body is directly proportional to the product of mass of the body and acceleration produced in the body by the action of the force, and it acts in the direction of acceleration.

The relation F ∝ m × a can be turned into an equation by putting in a constant k.

Thus,

$\large \: \sf \: F = k \times m \times a \: (where \: k \: is \: constant)$

The value of constant k in SI units is 1,so the above equation becomes

$\large \: \sf \: F = m \times a$

$\boxed{\red{\large\sf \: Force = mass \times acceleration}}$

Thus, Newton's 2nd law of motion gives us a relationship between Force & Acceleration.

Answer 2

Then, initial momentum of the object, t = mu

Final momentum of the object ,ct = mv

Change in momentum $= mu - mv= m(v - u)$

$The \: rate \: of \: change \: in momentum =(m(v - u))/t \\ According \: to \: Newton's \: second \: law \: of \: motion, \: we \: have \: F \infty \frac{m(v-u)}{t} \\ F=km (v-u) \\ F = kma \\ Here, \: a= v-u t \: = the \: rate \: of \: change \: of \: velocity. \: \\ = acceleration \: k \\ = a \: constant \: of \: proportionality \: Putting \: m n = 1 kg, \: \\ a = 1m * s ^ - 2 f \: becomes \: 1 N. \\ So, \: 1N = k * 1 kg * 1 * m * s ^ - 2 k = 1 \\ From \: equation (1), \: we \: have \: F = ma \\ (This \: represents \: the \: second \: law \: of \: motion.)$